Let $C_0(\mathbb{R})$ denote the analytic functions $f : \mathbb{R} \rightarrow \mathbb{R}$.

I wonder whether there a functions $f \in C_0(\mathbb{R})$ with $f \neq 0$, such that there is a constant $C$, with $$\left| \frac{d^kf}{d^kx} \right | \leq C$$ for all $k \geq 0$, and $\frac{d^kf}{dx}$ vanish both at $- \infty$ and $\infty$ for all $k \geq 0$.

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    $\begingroup$ What about $(\sin x)/x$? $\endgroup$ – Richard Stanley Oct 4 at 17:29
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    $\begingroup$ also, technically, since you didn't specify non-trivial, I guess $f\equiv 0$ is another exampl. $\endgroup$ – Willie Wong Oct 4 at 17:37
  • $\begingroup$ @Willie Wong Thanks for pointing it out, I amended it. $\endgroup$ – tobias Oct 4 at 17:54
  • $\begingroup$ @RichardStanley It seems $sin(x)/x$ is doing the job. However, is there an easy proof for it? $\endgroup$ – tobias Oct 4 at 21:10
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    $\begingroup$ @tobias: for a rescaled sinc function you can also get the proof using my answer. the fact that its Fourier transform is the rectangle function and not smooth is inconsequential: the Fourier transform of all its derivatives are bounded with compact support, so the decay at infinity can be also gotten via Riemann-Lebesgue. $\endgroup$ – Willie Wong Oct 5 at 1:55


Let $\phi$ be any smooth function with compact support on the interval $[-1,1]$.

Set $f$ to be the inverse Fourier transform of $\phi$.

Since $\phi$ is in Schwartz class, so is $f$, and all of its derivatives decay to zero as one approach $\pm\infty$.

You can estimate

$$ |f^{(k)}(x) | \lesssim \| |\xi|^k \phi(\xi) \|_{L^1} \leq 2 \|\phi\|_{L^\infty} =: C$$

$f$ is analytic by Paley-Wiener.

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    $\begingroup$ Why do you specify “inverse” Fourier transform and not simply the Fourier transform? For thepurposes of question it amounts to the same thing. $\endgroup$ – KConrad Oct 4 at 18:11
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    $\begingroup$ Because for my own sanity I like to keep track of which functions are on the frequency domain and which are on the time domain. (Force of habit.) $\endgroup$ – Willie Wong Oct 4 at 18:45

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